Regression: TypeScript no longer infers primitive types as widened types in conditional types.

[ghost]

TypeScript Version: 3.1.0-dev.20180918 <= affected version <= 3.4.0-dev.20190208

Search Terms:
Issues, PRs + Google: infer[ence] literal type
Searched the changelog of TS 3.1 to find out whether this might be an intended change.

Code

class Test<T> {
  makePromise<V>(value: T extends Promise<V> ? V : never): Promise<V> {
    return Promise.resolve(value);
  }
}

const test = new Test<Promise<number>>();
test.makePromise(5) // throws a compile-time error starting with [email protected]

Obs: infer V rather than <V> works across all versions in the above example.

Expected behavior:
In the above example V should be inferred to a primitive type (e.g. number in the given example but also applies to other primitive types)

Actual behavior:
Starting with [email protected], TypeScript infers the value in the above type to a literal type. The last working package is [email protected] (including 2.x and 3.x)

Playground Link:
https://www.typescriptlang.org/play/#src=class%20Test%3CT%3E%20%7B%0D%0A%20%20makePromise%3CV%3E(value%3A%20T%20extends%20Promise%3CV%3E%20%3F%20V%20%3A%20never)%3A%20Promise%3CV%3E%20%7B%0D%0A%20%20%20%20return%20Promise.resolve(value)%3B%0D%0A%20%20%7D%0D%0A%7D%0D%0A%0D%0Aconst%20test%20%3D%20new%20Test%3CPromise%3Cnumber%3E%3E()%3B%0D%0Atest.makePromise(5)%0D%0A

Related Issues:
none found

[RyanCavanaugh]

This is intended change.

The core problem here is that Promise's measured variance is backwards from what it "should" be - it appears to accept Ts but actually produces them. The user-side workaround is just to flip around the conditional type:

type Consumer<T> = {
  take(arg: T): void;
}

type Producer<T> = {
  produce(): T;
}

class Foo<T> {
  makeConsumer<U>(val: T extends Consumer<U> ? U : never): Consumer<U> {
    return {
      take(arg) {

      }
    }
  }

  // Fixed
  makeProducer<U>(val: Producer<U> extends T ? U : "other branch"): Producer<U> {
    return {
      produce() {
        return val as any;
      }
    }
  }
}

const f = new Foo<Consumer<number>>();
f.makeConsumer(10);

const g = new Foo<Producer<number>>();
g.makeProducer(10);

[ghost]

Sorry, unfortunately I didn't get it. You seem to suggest the following:

class Test<T> {
  // Switching around the conditional type as you propose:
  makePromise<V>(value: Promise<V> extends T ? V : never): Promise<V> {
    return Promise.resolve(value);
  }
}

This version will always enter the "other branch" if T subclasses Promise (as is intended obviously). Do you agree that this doesn't make sense?

@RyanCavanaugh: Could you please apply your solution to the original example so that it becomes clear how it should be rewritten? I think it would be great for others who stumble upon this ticket as well.

[typescript-bot]

This issue has been marked 'Working as Intended' and has seen no recent activity. It has been automatically closed for house-keeping purposes.